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Problem solution · Python

Minimum Time to Collect All Apples in a Tree

Minimum Time to Collect All Apples in a Tree: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
131 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Minimum Time to Collect All Apples in a Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 131 lines of Python from the credited upstream file minimum-time-to-collect-all-apples-in-a-tree.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Time to Collect All Apples in a Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(n) import collections  class Solution(object):    def minTime(self, n, edges, hasApple):        """        :type n: int        :type edges: List[List[int]]        :type hasApple: List[bool]        :rtype: int        """        graph = collections.defaultdict(list)        for u, v in edges:            graph[u].append(v)            graph[v].append(u)                result = [0, 0]        s = [(1, (-1, 0, result))]        while s:            step, params = s.pop()            if step == 1:                par, node, ret = params                ret[:] = [0, int(hasApple[node])]                for nei in reversed(graph[node]):                    if nei == par:                        continue                    new_ret = [0, 0]                    s.append((2, (new_ret, ret)))                    s.append((1, (node, nei, new_ret)))            else:                new_ret, ret = params                ret[0] += new_ret[0]+new_ret[1]                ret[1] |= bool(new_ret[0]+new_ret[1])        return 2*result[0]  # Time:  O(n)# Space: O(n)class Solution_Recu(object):    def minTime(self, n, edges, hasApple):        """        :type n: int        :type edges: List[List[int]]        :type hasApple: List[bool]        :rtype: int        """        def dfs(graph, par, node, hasApple):            result, extra = 0, int(hasApple[node])            for nei in graph[node]:                if nei == par:                    continue                count, found = dfs(graph, node, nei, hasApple)                result += count+found                extra |= bool(count+found)            return result, extra                graph = collections.defaultdict(list)        for u, v in edges:            graph[u].append(v)            graph[v].append(u)        return 2*dfs(graph, -1, 0, hasApple)[0]  # Time:  O(n)# Space: O(n)class Solution2(object):    def minTime(self, n, edges, hasApple):        """        :type n: int        :type edges: List[List[int]]        :type hasApple: List[bool]        :rtype: int        """        graph = collections.defaultdict(list)        for u, v in edges:            graph[u].append(v)            graph[v].append(u)                result = [0]        s = [(1, (-1, 0, result))]        while s:            step, params = s.pop()            if step == 1:                par, node, ret = params                tmp = [int(hasApple[node])]                s.append((3, (tmp, ret)))                for nei in reversed(graph[node]):                    if nei == par:                        continue                    new_ret = [0]                    s.append((2, (new_ret, tmp, ret)))                    s.append((1, (node, nei, new_ret)))            elif step == 2:                new_ret, tmp, ret = params                ret[0] += new_ret[0]                tmp[0] |= bool(new_ret[0])            else:                tmp, ret = params                ret[0] += tmp[0]        return 2*max(result[0]-1, 0)  # Time:  O(n)# Space: O(n)class Solution2_Recu(object):    def minTime(self, n, edges, hasApple):        """        :type n: int        :type edges: List[List[int]]        :type hasApple: List[bool]        :rtype: int        """        def dfs(graph, par, node, has_subtree):            result, extra = 0, int(hasApple[node])            for nei in graph[node]:                if nei == par:                    continue                count = dfs(graph, node, nei, hasApple)                result += count                extra |= bool(count)            return result+extra                graph = collections.defaultdict(list)        for u, v in edges:            graph[u].append(v)            graph[v].append(u)        return 2*max(dfs(graph, -1, 0, hasApple)-1, 0)

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